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How would you solve this:

3y-x^(2)+ln(xy) = 2

I am assuming you are asked to find dy/dx.

Firstly, I will do a little manipulation of the LHS:

3y-x^(2)+ln(xy) = 2

3y-x^(2)+ln x + lny = 2

Differentiating this on both sides wrt x gives

3 (dy/dx) -2x + 1/x +(1/y) (dy/dx) =0

( 3+ 1/y) (dy/dx) =2x -1/x

(3y + 1)/y * (dy/dx) = (2x^2 - 1) /x

dy/dx = [(2x^2 - 1)*y] /[ x * (3y+1)] (shown)

Hope this helps. Peace.

Calculus

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